## Calculating delta G, H, and S - 3 Ways

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Viraj B 3A
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Joined: Fri Sep 26, 2014 2:02 pm

### Calculating delta G, H, and S - 3 Ways

When we were discussing how to calculate the Gibbs Free Energy value for a reaction, we said that we can apply the 3 ways similar to when we were finding $\Delta$H.

1. Standard Reaction Enthalpies (or Gibbs Free Energy)
2. Hess's Law for Enthalpies (or Gibbs Free Energy)
3. Standard Enthalpies (or Gibbs Free Energy) of Formation.

I wanted to know whether these methods can be used to calculate the value of $\Delta$S? Since all three thermodynamic terms are state functions, can we apply the same methods used to calculate the $\Delta$H and $\Delta$G to calculate $\Delta$S?

martha-1I
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: Calculating delta G, H, and S - 3 Ways

I believe we can apply the same methods to calculate $\bigtriangleup$S as we do to find $\bigtriangleup$H as long as we use the $\bigtriangleup$S values instead of $\bigtriangleup$H. For example, when using the first method to find entropy for a balanced equation, we would need to find/obtain the $\bigtriangleup$S for each individual molecule in the equation(using Appendix 2A in the book) then use the equation $\bigtriangleup$S=$\sum \bigtriangleup S$products- $\sum \bigtriangleup S$reactants to calculate $\Delta S$.

Chem_Mod
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### Re: Calculating delta G, H, and S - 3 Ways

Methods 1) and 2) do apply to Entropy.

However, we don't have "Standard Entropies of Formation". Instead, the appendix will tabulate "Standard Molar Entropy" of substances. Note that these are denoted S, and NOT deltaS. Due to the Third Law of Thermodynamics, all substances have an absolute amount of entropy. Note that the Standard Molar Entropy of pure elements is NOT zero.

On the other hand, H and G can only be determined relatively, not absolutely, which is why we must arbitrarily set H and G of pure elements to zero.

In using the Standard Molar Entropies to find the entropy change for a reaction, though, nothing has changed. Just calculate "Final minus initial" as usual.